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What is the most distant object that can be seen from a particular location?

  1. Dec 23, 2015 #1
    Hello to everyone,

    I was wondering if somebody could help me with the specific issue I am interested in: What is the most distant object that can be seen from a particular location, "after" the horizon point?

    Here is the graphical preview of my issue.
    https://www.dropbox.com/s/5cxxmg570qttmfq/distance_to_horizon2.jpg?dl=0

    There is an observer point "O" located on shore, at some altitude h1. I can calculate the D1 distance to point H (the distance to the horizon), with the following formula:

    formula source:
    http://www-rohan.sdsu.edu/~aty/explain/atmos_refr/horizon.html
    (the upper formula assumes some theoretically ideal conditions: no impact of atmosphere layers, air temperature, wind and so on. It assumes the Earth to be a sphere, and takes into account the impact of light refraction also).

    My major concern is actually the D2 distance. What I do not understand is how to somehow predict this D2 distance for any location on Earth?

    For example, if Mount Everest (8.848 kilometers) the tallest peak on Earth would stand out from the sea, as a lonely island, a person standing on top of it, could have seen the horizon at the distance of 335 kilometers (D2 = sqrt (2*6371*7/6*8.848 + 8.848^2) = 335.8 km).
    So I guess in some theoretical imaginary case, D2 can not be larger than 335 kilometers.

    But what about other locations around the world, not near to Mount Everest?

    I understand that an answer to this question depends on the terrain of that location, it's altitude, and the height of the surrounding mountain peaks. Still does anyone know how can I somehow predict what the D2 distance should be (at least approximately)?

    Thank you for the reply in advance.

    Best regards,
    Bernard
     
    Last edited: Dec 23, 2015
  2. jcsd
  3. Dec 23, 2015 #2

    Bystander

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    Refraction "lifts" objects miles into the air on a fairly regular basis. Are you limiting the question?
     
  4. Dec 23, 2015 #3
    Thank you for the reply Bystander.
    The upper mentioned formula includes an effect of the "object lifting". It assumes it to be 1/7th of the Earth curvature correction.

    The problem is not with the formula, and I am not looking for a new one.

    What I am looking is a way to predict the "h2" variable. In some imaginal case I could just go with the mentioned Mount Everest height for h2. But this looks like an overestimate of the D2 distance for all other locations around the Earth.
     
  5. Dec 23, 2015 #4

    Bystander

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    Pike's Peak from Longmont --- zero height (or less) to in excess of foreground tree height. Half mile is to 50-100 feet as 100 miles is to "x"?
     
  6. Dec 23, 2015 #5
    Hi Bystander,
    I am not sure I understood you.
    You gave an example of being able to see Pike's Peak from Longmont in Colorado, both of which have the same altitude (elevation)?
    And then what does this mean: "50-100 feet as 100 miles is to "x"?"
     
  7. Dec 23, 2015 #6

    Bystander

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    "Pike's Peak:" a monolithic structure located in south-central Colorado standing 14,115 feet; Longmont, Colorado, elevation 5062 feet, (more like 4900 on NE corner) where I've a view due S to P-P, a difference of ~9000 feet. "Proportion:" a ratio; this is to that as something is to another thing.
    One half mile is to 50-100 feet as 100 miles is to "x," the difference in altitudes, or 9000 feet. Therefore, 90-180 times the altitude difference is a minimum for clear air. Given that Pike's Peak has been seen from Dodge City, ~300 miles, ....
     
  8. Dec 24, 2015 #7
    Thank you for the reply. But I am not sure I understand you.

    You are giving your personal example of how "Pike's Peak" (elevation: 14,115 feet) can be seen from "Longmont" (elevation: 5,062 feet). The distance between Pike's Peak and Longmont is 100 miles.

    Then you are mentioning the ratio:
    0.5 miles / 50 feet = 100 miles / x
    x = 10,000 feet

    0.5 feet / 100 feet = 100 miles / x
    x = 20,000 feet

    What is the meaning of this ratio?
    And what is the meaning of this 10,000 to 20,000 feet (1.9 to 3.9 miles)?
     
  9. Dec 24, 2015 #8

    jim mcnamara

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    I think you should tell us what you are trying to solve - ultimately. NOT what calculation you are trying to complete. What you are asking amounts to the assumption you need to accomplish something. You want help. Note: You are at the defining the problem step, not the solution step.

    Bystander's answers are correct, but you are clearly confused by them. You are trying to derive meaning from his numbers - which again returns us to assumptions and inferences -- not direct calculations.

    Beginnings are a wonderful place to start. So. One sentence would be great: Example: I want to describe everything visible on a given horizon so that I can do X with it. What is "X"?

    Also note carefully - your title as stated, only has a relative answer - that is: the view changes with the lat/long/altitude of each viewer. Which can change constantly. As stated there is no "general" answer.
     
  10. Dec 24, 2015 #9
    Thank you for the reply Jim.
    An issue of being able to see the most distant object from particular location is very common in hill climbing activities. So that's the "X" from your question.
    You will not find any other formula than mentioned one, or at least ones with a result far away from the mentioned one. All other formulas have all been derived from the assumption of Earth being a sphere and Pythagorean Theorem.

    I would be very grateful if you would explain Bystander's reply.
    As far as I understand it, he is giving me a specific example of how particular object (Pike's Peak) can be seen from a distance of 300 miles.
    If this is so, then this is not really what I have asked for.
     
  11. Dec 24, 2015 #10

    jim mcnamara

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    Well, I am with Bystander. I do not see what you asked for. Do you want an arbitrary value? That is what Pikes Peak and Mt Everest are - arbitrary.
    Do you want some kind of number - like 4500 feet higher elevation than the observer?

    The fact that you have the correct formula does not help any of us, if we do not apply values to it that are meaningful. I am an overeducated computer programmer in common parlance. We have a saying - GIGO. Garbage in garbage out. You want to avoid that.

    Based on your answer you have a tough problem, there is no single value for elevation for a hill climber. And the elevation has to be above ground level, the way pilots use elevation or altitude (AGL). In order to make calculations you will need input from a GPS receiver for the "hill climber" x,y,z --(lat/long/elevation), autocorrelation of that data to topo maps to get elevation above ground level, and then you have to mess around with the fact that you cannot see in all directions, except at the very top of the hill. Garmin GPS for example, does the xyz -> map position autocorrelation, google also has the google maps API you can use if you want. I would use the bluetooth interface available with some Garmin wrist models. Cellphone geo-accuracy is bad in the backcountry. Or non-existent. So cellphones are probably out of the picutre.

    Feed your XYZ and map topodata (like a GPX file ) into your app on a tablet or phone out on the trail. To get what you want using a single fudged value is not going to be helpful. People on Mt Washington vs folks on Mt Hood are going to want reasonable and different answers answers.

    FWIW what you are trying to solve is something that aeronautical navigation software deals with - like avoiding mountaintop impacts in bad weather. You have the reverse problem - you want to get a limit on visibility of objects.

    So, IMO, and others can please add more - making a single one-off calculation based on some fudged number is GIGO. Period. You can do it. But you seem to want real results. See this link on Enrico Fermi's fudging it method.
    [/PLAIN] [Broken]
    http://lesswrong.com/lw/7ij/rough_calculations_fermi_and_the_art_of_guessing/
     
    Last edited by a moderator: May 7, 2017
  12. Dec 25, 2015 #11
    I live by the water, sea level +3ft, on the top floor, 100ft up. Bangkok is 30 miles across the water at sea level +3ft.. The tallest building there is 1,030ft. Could I ever see it given clear air?

    Would it help if I went up to the roof, 110ft?

    One day I saw, with 8x binoculars, what looked like a building, a low rectangle, on the far shore, behind the setting sun. How far away could that have been? This is the furthest I have ever been able to see based upon the angle to my building.

    Is there ever clear air in SE Asia? Does the temperature and high humidity affect visibility?

    Thanks,
     
    Last edited: Dec 25, 2015
  13. Dec 26, 2015 #12

    jim mcnamara

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    Check out fata morgana - for one reasonable answer to your 'rectangular building'.
    Anytime you are in a densely populated area, there are incredible amounts of particulates in the air, these impede visibility.
    Inversion layers can create mirages like fata morgana. Sound also can travel further than you would expect : https://en.wikipedia.org/wiki/Inversion_(meteorology)
     
  14. Dec 31, 2015 #13

    CWatters

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    Perhaps i missed something but why can't you calculate d2 the same way you did d1?
     
  15. Jan 3, 2016 #14
    The Andromeda Galaxy, M31, at about 2.5 million light years away, is about as far away as you can normally see with the naked eye, unless you live in the lower half of the Southern hemisphere.

    https://en.wikipedia.org/wiki/Andromeda_Galaxy

    Some sharp-eyed people might be able to see M33 with perfectly dark skies, at about 3 million light years.

    https://en.wikipedia.org/wiki/Triangulum_Galaxy

    M81 is possible to observe, but just barely.

    https://en.wikipedia.org/wiki/Messier_81

    Oh, did you mean objects on Earth? Never mind ...
     
  16. Mar 31, 2016 #15
    Also, remember that Earth is not a perfect sphere. Although Everest is at a higher altitude from sea level there are other objects that ar further from the center of the Earth. I believe Mauna Koa in Hawaii may be. And as I believe someone already mentioned, the atmosphere can also distort distances and heights. I was driving to Calgary, Alberta once very early in the morning and normally I was still too far away from the city to be able to see its lights. I lived in a small town about 40 minutes drive from Calgary. However, that morning I not only could already see the city lights from a place where they should not normally have been visible, but they were also upside down. There lay the city-scape of Calgary, Alberta laid out for me in the sky completely upside-down. It was one of the coolest things I have ever seen. And, no, I was only on the effects of one cup of coffee.
     
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